3.15.62 \(\int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} (1+(-1-x) \log ^2(x)))}{\log (5) \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {5+e^{x-\left (e^{4+x+\frac {1}{\log (x)}}-\frac {3}{x}\right ) x}}{\log (5)} \]

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Rubi [F]  time = 1.49, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log (5) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 + x - E^((1 + (4 + x)*Log[x])/Log[x])*x)*(Log[x]^2 + E^((1 + (4 + x)*Log[x])/Log[x])*(1 + (-1 - x)*L
og[x]^2)))/(Log[5]*Log[x]^2),x]

[Out]

Defer[Int][E^(3 + x - E^(4 + x + Log[x]^(-1))*x), x]/Log[5] - Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/
Log[x])*x + Log[x]^(-1)), x]/Log[5] - Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/Log[x])*x + Log[x]^(-1))
*x, x]/Log[5] + Defer[Int][E^(7 + 2*x - E^((1 + (4 + x)*Log[x])/Log[x])*x + Log[x]^(-1))/Log[x]^2, x]/Log[5]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \left (\log ^2(x)+e^{\frac {1+(4+x) \log (x)}{\log (x)}} \left (1+(-1-x) \log ^2(x)\right )\right )}{\log ^2(x)} \, dx}{\log (5)}\\ &=\frac {\int \left (e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x}-\frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \left (-1+\log ^2(x)+x \log ^2(x)\right )}{\log ^2(x)}\right ) \, dx}{\log (5)}\\ &=\frac {\int e^{3+x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \left (-1+\log ^2(x)+x \log ^2(x)\right )}{\log ^2(x)} \, dx}{\log (5)}\\ &=\frac {\int e^{3+x-e^{4+x+\frac {1}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \left (\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )+\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) x-\frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )}{\log ^2(x)}\right ) \, dx}{\log (5)}\\ &=\frac {\int e^{3+x-e^{4+x+\frac {1}{\log (x)}} x} \, dx}{\log (5)}-\frac {\int \exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) \, dx}{\log (5)}-\frac {\int \exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right ) x \, dx}{\log (5)}+\frac {\int \frac {\exp \left (7+2 x-e^{\frac {1+(4+x) \log (x)}{\log (x)}} x+\frac {1}{\log (x)}\right )}{\log ^2(x)} \, dx}{\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.07, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{3+x-e^{4+x+\frac {1}{\log (x)}} x}}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + x - E^((1 + (4 + x)*Log[x])/Log[x])*x)*(Log[x]^2 + E^((1 + (4 + x)*Log[x])/Log[x])*(1 + (-1
- x)*Log[x]^2)))/(Log[5]*Log[x]^2),x]

[Out]

E^(3 + x - E^(4 + x + Log[x]^(-1))*x)/Log[5]

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fricas [A]  time = 0.94, size = 26, normalized size = 0.90 \begin {gather*} \frac {e^{\left (-x e^{\left (\frac {{\left (x + 4\right )} \log \relax (x) + 1}{\log \relax (x)}\right )} + x + 3\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="fricas")

[Out]

e^(-x*e^(((x + 4)*log(x) + 1)/log(x)) + x + 3)/log(5)

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giac [A]  time = 0.70, size = 20, normalized size = 0.69 \begin {gather*} \frac {e^{\left (-x e^{\left (x + \frac {1}{\log \relax (x)} + 4\right )} + x + 3\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="giac")

[Out]

e^(-x*e^(x + 1/log(x) + 4) + x + 3)/log(5)

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maple [A]  time = 0.08, size = 29, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-x \,{\mathrm e}^{\frac {x \ln \relax (x )+4 \ln \relax (x )+1}{\ln \relax (x )}}+3+x}}{\ln \relax (5)}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x-1)*ln(x)^2+1)*exp(((4+x)*ln(x)+1)/ln(x))+ln(x)^2)*exp(-x*exp(((4+x)*ln(x)+1)/ln(x))+3+x)/ln(5)/ln(x)
^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(5)*exp(-x*exp((x*ln(x)+4*ln(x)+1)/ln(x))+3+x)

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maxima [A]  time = 0.71, size = 20, normalized size = 0.69 \begin {gather*} \frac {e^{\left (-x e^{\left (x + \frac {1}{\log \relax (x)} + 4\right )} + x + 3\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*log(x)^2+1)*exp(((4+x)*log(x)+1)/log(x))+log(x)^2)*exp(-x*exp(((4+x)*log(x)+1)/log(x))+3+x)
/log(5)/log(x)^2,x, algorithm="maxima")

[Out]

e^(-x*e^(x + 1/log(x) + 4) + x + 3)/log(5)

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mupad [B]  time = 1.08, size = 22, normalized size = 0.76 \begin {gather*} \frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^4\,{\mathrm {e}}^{\frac {1}{\ln \relax (x)}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^3\,{\mathrm {e}}^x}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - x*exp((log(x)*(x + 4) + 1)/log(x)) + 3)*(log(x)^2 - exp((log(x)*(x + 4) + 1)/log(x))*(log(x)^2*(x
+ 1) - 1)))/(log(5)*log(x)^2),x)

[Out]

(exp(-x*exp(4)*exp(1/log(x))*exp(x))*exp(3)*exp(x))/log(5)

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sympy [A]  time = 2.53, size = 22, normalized size = 0.76 \begin {gather*} \frac {e^{- x e^{\frac {\left (x + 4\right ) \log {\relax (x )} + 1}{\log {\relax (x )}}} + x + 3}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*ln(x)**2+1)*exp(((4+x)*ln(x)+1)/ln(x))+ln(x)**2)*exp(-x*exp(((4+x)*ln(x)+1)/ln(x))+3+x)/ln(
5)/ln(x)**2,x)

[Out]

exp(-x*exp(((x + 4)*log(x) + 1)/log(x)) + x + 3)/log(5)

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